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The \(L^{\mathrm{M}}\) and \(L^{\mathrm{N}}\) alleles at the MN blood-group locus exhibit codominance. Give the expected genotypes and phenotypes and their ratios in progeny resulting from the following crosses. a. \(L^{\mathrm{M}} L^{\mathrm{M}} \times L^{\mathrm{M}} L^{\mathrm{N}}\) \(\mathbf{b .} L^{\mathrm{N}} L^{\mathrm{N}} \times L^{\mathrm{N}} L^{\mathrm{N}}\) \(\mathbf{c .} L^{\mathbf{M}} L^{\mathbf{N}} \times L^{\mathbf{M}} L^{\mathbf{N}}\) \(\mathbf{d .} L^{\mathrm{M}} L^{\mathrm{N}} \times L^{\mathrm{N}} L^{\mathrm{N}}\) \(\mathbf{e} . L^{\mathrm{M}} L^{\mathrm{M}} \times L^{\mathrm{N}} L^{\mathrm{N}}\)

Short Answer

Expert verified
a. Genotypes 1:1 (M:MN), phenotypes 50% M, 50% MN. b. Genotype/phenotype 100% N. c. Genotypes 1:2:1, phenotypes 25% M, 50% MN, 25% N. d. Genotypes 1:1, phenotypes 50% MN, 50% N. e. Genotype/phenotype 100% MN.

Step by step solution

01

Understand Codominance

In codominance, both alleles contribute equally to the phenotype. In the case of the MN blood group, the alleles \(L^M\) and \(L^N\) are codominant, meaning both express when present together.
02

Analyze Cross (a)

For the cross \(L^{M} L^{M} \times L^{M} L^{N}\), the possible genotypes of the offspring are:- \(L^M L^M\) (homozygous MM)- \(L^M L^N\) (heterozygous MN)The expected genotype ratio is 1:1. The phenotypes will be 50% M and 50% MN, showing the genotype ratio.
03

Analyze Cross (b)

For the cross \(L^{N} L^{N} \times L^{N} L^{N}\), both parents are homozygous NN, so all offspring will have the genotype \(L^N L^N\). Thus, all phenotypes will be N, with a ratio of 100% NN.
04

Analyze Cross (c)

For the cross \(L^{M} L^{N} \times L^{M} L^{N}\), the possible genotypes of the offspring are:- \(L^M L^M\) (homozygous MM)- \(L^M L^N\) (heterozygous MN)- \(L^N L^N\) (homozygous NN)The expected genotype ratio is 1:2:1. Therefore, the phenotypic ratio will be:- 25% M (\(L^M L^M\))- 50% MN (\(L^M L^N\))- 25% N (\(L^N L^N\))
05

Analyze Cross (d)

For the cross \(L^{M} L^{N} \times L^{N} L^{N}\), the possible genotypes of the offspring are:- \(L^M L^N\) (heterozygous MN)- \(L^N L^N\) (homozygous NN)The expected genotype ratio is 1:1. The phenotypic ratio will be 50% MN and 50% N.
06

Analyze Cross (e)

For the cross \(L^{M} L^{M} \times L^{N} L^{N}\), the possible genotypes of the offspring are:- \(L^M L^N\) (heterozygous MN)All offspring will thus have the phenotype MN, with a ratio of 100% MN.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

MN Blood Group
The MN blood group is a fascinating aspect of genetic study because it showcases the concept of codominance. Unlike other types of genetic inheritance, codominance occurs when two different alleles share equal dominance, and both contribute to the phenotype.

In humans, the MN blood group is determined by two alleles: \(L^M\) and \(L^N\). Both alleles are found at the MN blood group locus on chromosome 4. People can have three possible combinations of these alleles, resulting in three distinct phenotypes:
  • \(L^M L^M\) - The M phenotype.
  • \(L^N L^N\) - The N phenotype.
  • \(L^M L^N\) - The MN phenotype, where both M and N antigens are expressed equally.
Understanding the MN blood group helps us appreciate how different inheritance patterns contribute to genetic diversity among populations.
Genotype and Phenotype Ratios
When studying genetics, it's important to understand the difference between genotypes and phenotypes. Genotype refers to the genetic makeup of an organism—in the case of the MN blood group, it would be the combination of \(L^M\) and \(L^N\) alleles. Phenotype, on the other hand, is the observable trait or expression resulting from the genotype.

Predicting the expected genotype and phenotype ratios in offspring is a core part of genetics. By analyzing the specific genetic crosses:
  • For a cross like \(L^M L^M \times L^M L^N\), you can expect offspring genotypes of \(L^M L^M\) and \(L^M L^N\), leading to a phenotype ratio of 50% M and 50% MN.
  • With a cross \(L^N L^N \times L^N L^N\), all offspring will be \(L^N L^N\), meaning a 100% N phenotype.
  • In a more varied cross like \(L^M L^N \times L^M L^N\), the genotype ratio becomes 1:2:1, with a phenotypic result of 25% M, 50% MN, and 25% N.
These ratios help understand how traits are passed on and expressed in a population.
Alleles
Alleles are alternative forms of a gene found at the same place on a chromosome. For the MN blood group, \(L^M\) and \(L^N\) are different alleles.

Each person inherits two alleles for each gene: one from their mother and one from their father. This pair of alleles will determine their genetic characteristics. Depending on the combination of these alleles, a person can exhibit different traits.

Understanding alleles is crucial in predicting genetic outcomes and ratios. Some genes show dominance, where one allele completely masks the presence of another, but in codominance, such as with the MN alleles, both alleles are equally expressed. This leads to a unique phenotype that exhibits features of both alleles, like the MN phenotype that shows both M and N blood group antigens.
Inheritance Patterns
Inheritance patterns describe how genetic traits are transmitted from parents to offspring. The MN blood group highlights codominance as a key pattern in genetics.

In codominance, the offspring receives one allele from each parent, and both alleles are expressed in the organism. This differs from complete dominance, where only one allele's traits are observable.
  • For example, if one parent has the genotype \(L^M L^N\), and the other parent has \(L^N L^N\), offspring can inherit either \(L^M L^N\) or \(L^N L^N\), manifesting in MN or N phenotypes accordingly.
By analyzing potential crosses, you can predict the inheritance and expression of genes, aiding in genetic counseling and understanding hereditary conditions.

Such knowledge helps build a clear picture of how diverse traits persist in populations, leading to greater genetic variety and adaptation over time.

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Most popular questions from this chapter

When a Chinese hamster with white spots is crossed with another hamster that has no spots, approximately 12 of the offspring have white spots and 12 have no spots. When two hamsters with white spots are crossed, 23 of the offspring possess white spots and 13 have no spots. a. What is the genetic basis of white spotting in Chinese hamsters? b. How might you go about producing Chinese hamsters that breed true for white spotting?

Long ears in some dogs are an autosomal dominant trait. Two dogs mate and produce a litter in which \(75 \%\) of the puppies have long ears. Of the dogs with long ears in this litter, 13 are known to be phenocopies. What are the most likely genotypes of the two parents of this litter?

A woman has blood-type A M. She has a child with blood-type AB MN. Which of the following blood types could not be that of the child's father? Explain your reasoning. \(\begin{array}{lll}\text { George } & \mathrm{O} & \mathrm{N} \\ \text { Tom } & \mathrm{AB} & \mathrm{MN} \\ \text { Bill } & \mathrm{B} & \mathrm{MN} \\\ \text { Claude } & \mathrm{A} & \mathrm{N} \\ \text { Henry } & \mathrm{AB} & \mathrm{M}\end{array}\)

A variety of opium poppy (Papaver somniferum L.) with lacerate leaves was crossed with a variety that has normal leaves. All the \(\mathrm{F}_{1}\) had lacerate leaves. Two \(\mathrm{F}_{1}\) plants were interbred to produce the \(\mathrm{F}_{2}\). Of the \(\mathrm{F}_{2}, 249\) had lacerate leaves and 16 had normal leaves. Give genotypes for all the plants in the \(\mathrm{P}, \mathrm{F}_{1},\) and \(\mathrm{F}_{2}\) generations. Explain how lacerate leaves are determined in the opium poppy.

Suppose that you are tending a mouse colony at a genetic research institute, and one day you discover a mouse with twisted ears. You breed this mouse with twisted ears and find that the trait is inherited. Both male and female mice may have twisted ears, but when you cross a twisted-eared male with a normal- eared female, you obtain results that differ from those obtained when you cross a twisted-eared female with a normal-eared male: the reciprocal crosses give different results. Describe how you would determine whether this trait results from a sexlinked gene, a sex-influenced gene, genetic maternal effect, a cytoplasmically inherited gene, or genomic imprinting. What crosses would you conduct, and what results would be expected with these different types of inheritance?

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